General Chemistry I: Chemical Reactions and Problem Solving (Lecture 9, Chapter 4)

Chemical Reactions and Problem Solving

Introduction to Chemical Reactions

Chemical reactions are processes in which substances (reactants) are transformed into new substances (products). Understanding how to represent, balance, and analyze chemical reactions is fundamental in general chemistry.

• Chemical Equation: A symbolic representation of a chemical reaction, showing reactants and products.

• Reactants: Substances present before the reaction (left side of the equation).

• Products: Substances formed as a result of the reaction (right side of the equation).

• Example: $N_2(g) + 3H_2(g) ightarrow 2NH_3(g)$

Balancing Chemical Equations

Balanced chemical equations have the same number of each type of atom on both sides. Balancing is achieved by adjusting coefficients, not subscripts.

• Steps to Balance:

  1. Write the skeletal equation.

  2. Adjust coefficients to balance atoms.

  3. Use fractional coefficients if necessary, but final coefficients must be whole numbers.

  4. Reduce coefficients to the smallest whole-number ratio.

• Example:

  • Unbalanced: $H_2 + O_2 ightarrow H_2O$

  • Balanced (fractional): $H_2 + rac{1}{2}O_2 ightarrow H_2O$

  • Balanced (whole): $2H_2 + O_2 ightarrow 2H_2O$

• Important: Never change subscripts when balancing equations.

Stoichiometry

Reaction Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction, based on the balanced chemical equation.

• Mole Ratios: Coefficients in the balanced equation indicate the ratio of moles of each substance.

• Example: $2H_2(g) + O_2(g) ightarrow 2H_2O(g)$

  • 2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$.

• Conversions:

  • Use molar mass to convert between grams and moles.

  • Use Avogadro's number ($6.022 imes 10^{23}$) to convert between moles and number of particles.

Stoichiometric Calculations

Stoichiometric calculations allow prediction of the amount of product formed or reactant required in a chemical reaction.

• General Steps:

  1. Convert given mass to moles using molar mass.

  2. Use mole ratio from the balanced equation to find moles of desired substance.

  3. Convert moles back to mass if required.

• Example Problem:

  • Given: 10 g Ni ($M_{Ni} = 58.69$ g/mol)

  • Find: Number of moles and atoms of Ni

  • Moles: $10 ext{ g Ni} imes rac{1 ext{ mol}}{58.69 ext{ g}} = 0.17 ext{ mol}$

  • Atoms: $0.17 ext{ mol} imes 6.022 imes 10^{23} ext{ atoms/mol} = 1 imes 10^{23} ext{ atoms}$

Percent Composition and Empirical Calculations

Percent composition refers to the percentage by mass of each element in a compound.

• Formula: $ext{Percent } X = rac{ ext{mass of } X ext{ in compound}}{ ext{molar mass of compound}} imes 100$

• Example: For $K_2SO_4$ ($M = 174.26$ g/mol), mass of K in 5 g $K_2SO_4$:

  • Percent K: $rac{2 imes 39.10}{174.26} imes 100 = 44.9 ext{ ext{(rounded)}}$

  • To find mass of $K_2SO_4$ containing 5 g K: $5 ext{ g K} imes rac{100 ext{ g } K_2SO_4}{44.9 ext{ g K}} = 11.1 ext{ g } K_2SO_4$

Density and Mole Calculations

Density is used to convert between mass and volume, which can then be related to moles and atoms.

• Formula: $d = rac{m}{V}$

• Example: For $Br_2$ (density = 3.1 g/mL), 5 mL sample:

  • Mass: $3.1 ext{ g/mL} imes 5 ext{ mL} = 15.5 ext{ g}$

  • Moles: $15.5 ext{ g} imes rac{1 ext{ mol}}{159.81 ext{ g}} = 0.097 ext{ mol}$

  • Atoms: $0.097 ext{ mol} imes 2 imes 6.022 imes 10^{23} = 1.2 imes 10^{23} ext{ Br atoms}$

Average Atomic Mass

The average atomic mass of an element is calculated using the masses and relative abundances of its isotopes.

• Formula: $ext{Average Atomic Mass} = extstyleigg( ext{Isotope}_1 ext{ Mass} imes ext{Abundance}_1igg) + igg( ext{Isotope}_2 ext{ Mass} imes ext{Abundance}_2igg)$

• Example: Bromine isotopes:

  • $^{79}$Br: 78.92 amu (50.69%)

  • $^{81}$Br: 80.92 amu (49.31%)

• Calculation: $78.92 imes 0.5069 + 80.92 imes 0.4931 = 79.91$ amu

Chemical Nomenclature

Naming Compounds

Chemical nomenclature is the system for naming chemical substances. The rules differ for ionic, covalent, and acids.

• Ionic Compounds: Name the cation first, then the anion. Use Roman numerals for transition metals.

• Covalent Compounds: Use prefixes (mono-, di-, tri-, etc.) to indicate the number of atoms.

• Acids:

  • If the anion ends in -ide, the acid name begins with 'hydro-' and ends with '-ic acid' (e.g., HBr: hydrobromic acid).

  • If the anion ends in -ate, the acid name ends with '-ic acid' (e.g., HNO3: nitric acid).

• Examples:

  • Mg(OH)2: magnesium hydroxide

  • FeSO4: iron(II) sulfate

  • P2O3: diphosphorus trioxide

  • SO2: sulfur dioxide

Limiting Reactants and Percent Yield

Limiting and Excess Reactants

In reactions with multiple reactants, the limiting reactant is the one that is completely consumed first, thus limiting the amount of product formed. The excess reactant is left over after the reaction.

• Identifying Limiting Reactant:

  1. Calculate the amount of product formed from each reactant; the smallest amount is the theoretical yield.

  2. Alternatively, compare the given molar ratio to the required ratio from the balanced equation.

• Theoretical Yield: The maximum amount of product that can be formed from the limiting reactant.

• Example: $2H_2 + O_2 ightarrow 2H_2O$

  • Given: 42 mol $H_2$, 12 mol $O_2$

  • Limiting reactant: $O_2$ (since $42/2 = 21$ and $12/1 = 12$; $O_2$ produces less $H_2O$)

  • Theoretical yield: $12 imes 2 = 24$ mol $H_2O$

Percent Yield

Percent yield compares the actual yield (amount of product obtained) to the theoretical yield (maximum possible amount).

• Formula: $ext{Percent Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100$

• Example: If 7.0 g $N_2$ is obtained but 8.0 g was possible, percent yield is $rac{7.0}{8.0} imes 100 = 87.5 ext{ ext{(rounded)}}$

Representative Stoichiometry Problems

Sample Calculations

Several example problems illustrate the application of stoichiometry, limiting reactant, and percent yield concepts.

• Combustion of Hydrocarbons:

  • Balanced equation for octane: $2C_8H_{18}(l) + 25O_2(g) ightarrow 16CO_2(g) + 18H_2O(g)$

  • Molar mass calculations: $C_8H_{18}$: $8 imes 12.01 + 18 imes 1.01 = 114.26$ g/mol; $CO_2$: $12.01 + 2 imes 16.00 = 44.01$ g/mol

  • Conversion factors: Use mole ratios and molar masses to estimate mass of $CO_2$ produced from a given mass of $C_8H_{18}$.

• Photosynthesis:

  • $6CO_2(g) + 6H_2O(l) ightarrow C_6H_{12}O_6(s) + 6O_2(g)$

  • Given mass of $CO_2$, calculate mass of glucose produced using molar masses and mole ratios.

• Acid Rain Formation:

  • $2SO_2(g) + O_2(g) + 2H_2O(l) ightarrow 2H_2SO_4(aq)$

  • Given mass of $SO_2$, calculate mass of $H_2SO_4$ formed.

• Neutralization Reaction:

  • $Mg(OH)_2(s) + 2HCl(aq) ightarrow 2H_2O(l) + MgCl_2(aq)$

  • Given mass of $Mg(OH)_2$, calculate mass of $HCl$ neutralized.

Tables

Stoichiometric Conversion Table

This table summarizes the steps for converting between mass, moles, and number of particles in stoichiometric calculations.

Given	Conversion	Use
Mass of A	Divide by molar mass of A	Get moles of A
Moles of A	Use coefficients from balanced equation	Get moles of B
Moles of B	Multiply by molar mass of B	Get mass of B
Moles of substance	Multiply by Avogadro's number	Get number of particles

Sample Isotope Table (Bromine)

Isotope	Mass (amu)	Abundance (%)
$^{79}$Br	78.92	50.69
$^{81}$Br	80.92	49.31

Summary

This guide covers the essential concepts of chemical reactions, stoichiometry, chemical nomenclature, limiting reactants, and percent yield, with representative problems and tables to aid understanding and exam preparation.

Additional info: Some context and explanations have been expanded for clarity and completeness.